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# Find maximum of a discontinuous function.

## Find maximum of a discontinuous function.

### RE: Find maximum of a discontinuous function.

You assigned a function to a single-valued variable, which is what's causing the error. You need to make x an indexed variable, and make Taa and indexed variable.

### RE: Find maximum of a discontinuous function.

(OP)
Sorry, I didn't got it.
I tried, but nothing.
I don't know how to do that.

### RE: Find maximum of a discontinuous function.

I fixed a logic error in your first compare, where you used 0 < x ≤ 1 instead of 0 ≤ x ≤ 1, since x is allowed to be equal to zero, resulting in an undefined output. As for what you are supposedly attempting to do, it doesn't make sense:

http://files.engineering.com/getfile.aspx?folder=3...

### RE: Find maximum of a discontinuous function.

The underlying problem with your error is that x is a range variable not a vector. In Mathcad, a range variable is an iterator (think of it as an implied for-loop) and you must (normally) have an instance of a range variable on the left hand side of an assignment as well as on the right. Were Mathcad to have executed your statement Taa:=Ta(x), you would probably just have ended up with the value of Ta for the last value of x.

Given IRStuff's amended worksheet, you can apply the modified Ta to x (which is now in vector form) by using the vectorization operator. If you don't use the vectorization operator, then Mathcad will try to apply Ta to the vector x rather than each element of x (which is what you want). Unfortunately, Mathcad is a bit inconsistent with the use of logical operators and vectors. It automatically vectorizes single comparisons but doesn't allow the upper and lower bounded form of comparison you use; nor does it allow the logical operators and, or, xor or not to directly apply to vectors (or any other type of array) - the vectorize operator must be used.

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